Question

If $\frac{tan3\theta -1}{tan3\theta +1}=\sqrt{3}$ , then the general value of $\theta$ is :

• A. $\frac{n\pi }{3}-\frac{\pi }{12}$
• B. $n\pi +\frac{7\pi }{12}$
• C. $\frac{n\pi }{3}+\frac{7\pi }{36}$
• D. $n\pi +\frac{\pi }{12}$

Answer 1

Answer: (C)

Given that, $\frac{tan3\theta -1}{tan3\theta +1}=\sqrt{3}$
$\Rightarrow tan3\theta -1=\sqrt{3}tan3\theta +\sqrt{3}$
$\Rightarrow tan3\theta -1-\sqrt{3}tan3\theta -\sqrt{3}=0$
$\Rightarrow tan3\theta \left(1-\sqrt{3}\right)-\left(1+\sqrt{3}\right)=0$
$\Rightarrow tan3\theta =\frac{1+\sqrt{3}}{1-\sqrt{3}}$
$\Rightarrow tan3\theta =tan105^{\circ }=tan\frac{7\pi }{12}$
$\therefore 3\theta =n\pi +\frac{7\pi }{12}$
$\Rightarrow \theta =\frac{n\pi }{3}+\frac{7\pi }{12}$